Question

[Bí Mật]

SO SÁNH :        \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\) và \(-\frac{1}{2}\)

help me!!!!!!!

Answer 1

[╰Nguyễn Trí Nghĩa (team toán học)ッ...]

Đặt \(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).........\left(\frac{1}{100^2}-1\right)\)

\(\Rightarrow A=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}............\frac{1-100^2}{100}\)

\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}............\frac{-9999}{100^2}\)

\(\Rightarrow A=\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}...............\frac{-99.101}{100^2}\)

\(\Rightarrow A=\frac{-\left(1.2.3.............99\right).\left(3.4.5............101\right)}{\left(2.3.4......100\right).\left(2.3.4.............100\right)}\)

\(\Rightarrow A=\frac{-1.101}{100.2}=\frac{-101}{200}\)

Vậy \(A=\frac{-101}{200}\)

Chúc bn học tốt

Answer 2

[IS]

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

=>\(\)

=>\(A< -\left(\frac{1.2.3....99}{2.3.4...100}\right)=-\frac{1}{100}\)

Mà \(-\frac{1}{100}>-\frac{1}{2}\)

=>\(A>-\frac{1}{2}\) đúng ko nhỉ